Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.6 - Complex Numbers - 1.6 Exercises - Page 64: 65



Work Step by Step

$x^{2}+3x+7=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For this equation, $a=1$, $b=3$ and $c=7$. Substitute the known values in the formula: $x=\dfrac{-3\pm\sqrt{3^{2}-4(1)(7)}}{2(1)}=\dfrac{-3\pm\sqrt{9-28}}{2}=\dfrac{-3\pm\sqrt{-19}}{2}=...$ $...=\dfrac{-3\pm\sqrt{19}i}{2}=-\dfrac{3}{2}\pm\dfrac{\sqrt{19}}{2}i$ The answer is $x=-\dfrac{3}{2}\pm\dfrac{\sqrt{19}}{2}i$
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